Grrr. Typos. Getting sleepy. 
Nut you jot th idea
I’m sorry but that’s not correct.
Measure distance in light years and time in years.
V = d / t
Say the distance is 1 light year and the time is 1 year.
V = (d / t) = (1 year / 1 year) = 1 = c.
Measure distance in meters and time in meters of light travel time.
Say the distance is 3x10[sup] 8[/sup] meters and the time 3x10[sup] 8[/sup] meters of light travel time.
V = d / t = (3x10[sup] 8[/sup] m) / (3x10[sup]8[/sup]m) = 1 = c.
If you don’t believe this works check any introductory SR text. “Spacetime Physics” by Wheeler is a good one. But, regardless, I’m done arguing this point. Physicists use this convention every day.
Believe me I don’t need to consult an introductory text here.
Incorrect Ring
d = light-year “light-year” is a unit of distance
t = year “year” is a unit of time (year is not a unit of distance
light-year does not equal year anymore than a kilowatt-hour is equal to a kilowatt
v = light-year/year
3x10[sup]8[/sup] m is not a unit of time.
Perhaps you should review the end of that other thread when the equation of special relativity space-time intervals is discussed. You cannot blindly exchange distance and time. They are integral, but not interchangeable. We can review that discussion again also if you wish.
Let’s go back to my P.S. post. That was really the more important flaw in your use of that equation. It seems that you have taken an equation out of its rightful context and thus derived an erroneous conclusion.
Quoting from Fundamentals of Physics Extended Fifth Edition by Halliday/Resnick/Walker (bolding is mine)
I believe this was the equation you were deriving from.
Notice that is says “particle” and not system. There is a very good reason for this. This equation is not valid for systems of multiple particles.
Let’s look back at the primary total energy equation from which this equation was derived:
E[sub]t[/sub][sup]2[/sup] = E[sub]k[/sub][sup]2[/sup] + E[sub]r[/sub][sup]2[/sup]
“The square of the total energy of a system is equal to the sum of the squares of the total kinetic energy of the system and the rest energy of the system”
So the equation you derived from is only valid in systems for which E[sub]k[/sub] = pc. That is very unlikely to be the case in a system of more than one particle.
Undead Dude, I’m sorry, but you’re absolutely 100% wrong about needing c in your equations. Units themselves are only a convenience; we’re used to measuring distance and time in different units, but there is absolutely no fundamental reason that this must be done. It’s a convenient choice, no more. I can work in whatever system of units I want, as long as I’m consistent, and if I decide I want to work in a system of units where c=1, that’s perfectly acceptable. Some people get confused when one does this, but there’s nothing wrong with it whatsoever.
Think of it in analogy with E&M; in SI units, the force equation is q[sub]1[/sub]q[sub]2[/sub]/4[symbol]pe[/symbol][sub]0[/sub]r[sup]2[/sup], in Gaussian units, it’s just q[sub]1[/sub]q[sub]2[/sub]/r[sup]2[/sup], in Heaviside-Lorentz, it’s q[sub]1[/sub]q[sub]2[/sub]/4[symbol]p[/symbol]r[sup]2[/sup]. All are equally valid equations for force, it’s just that we’re measuring charge differently in each system of units; the units are something we get to pick, not something nature tells us.
As far as conservation of momentum goes, Ring is 100% correct; the total momentum 4-vector is conserved in every process. You can see that p[sup][symbol]m[/symbol][/sup] must be conserved because at each vertex in the diagrams that describe the process, there’s a [symbol]d[/symbol] function that enforces it; ergo, it’s conserved for the process as a whole; the net 4 momentum going in is the same as the net 4 momentum going out.
One implication is that the mass of the system in total is also conserved, where the “mass” of the system in total just means the length of the total momentum four vector. What it doesn’t mean is that the mass of the system in total is the sum of the masses of the individual components. The equation m[sup]2[/sup] = E[sup]2[/sup]-p[sup]2[/sup] holds regardless of whether it’s describing one particle, or three, or 107. It just defines the invariant length of p[sup][symbol]m[/symbol][/sup].
No text books handy to cite this for you, but any book on quantum field theory, for example, would work as a reference, even such a piece of garbage as Peskin.
Here, I’ll make this concrete. Suppose I have an electron-positron pair, in the center of momentum frame. Then the electron has p[symbol][sup]m[/sup][/symbol] = (E,p) and the positron has p[symbol][sup]m[/sup][/symbol] = (E,-p). Each has mass equal to the electron mass; you’d use this to find E given p, via E[sup]2[/sup] = m[sup]2[/sup] + p[sup]2[/sup]. The total net vector is then P[sup][symbol]m[/symbol][/sup] = (2E,0). It has length P[sup][symbol]m[/symbol][/sup] P[sub][symbol]m[/symbol][/sub] = M[sup]2[/sup] = 4E[sup]2[/sup], where now this defines M.
If they hit each other and produce a pair of photons, the photons go off each with some momentum q. We can find out what the norm of q is by using that the total momentum is conserved; this means that each photon has energy E, and hence q[sup]2[/sup] = E[sup]2[/sup]. The “mass” of the system, M, is the same before and after, although before we had an electron-positron pair (massive) and after we had a pair of photons (massless). I think this is what Ring means.
Now, granted, I would call the mass at the beginning 2m[sub]e[/sub] and the mass at the end 0, but if you’re using mass to mean the length of the total momentum, it’s conserved, period, end of story. [nerd hat off]
“But, regardless, I’m done arguing this point.” – Ring
Okay, then please argue the other point. The point of the invalid context of that equation.
And if you are so certain of these points, you are free to cite superior references.
Okay, lets drop the whole unit issue for the moment. I think we have a mess there that is hard to untangle. I am quite aware of the convention of manipulating units for convenience, but manipulating units is something that is far from arbitrary. For now, lets look at the rest of the picture.
g8rguy, I thought it was pretty clear that we were not using mass here to describe the magnitude of a momentum 4 vector but the “invariant mass” or “rest mass”.
Let’s consider another scenario-- 2 anti-parallel objects moving at high speed. No antimatter. They don’t need to be subatomic particles. For now, I’m not discussing a collision.
What is the kinetic energy of this system? According to your equation above it should be equal to p[sup]2[/sup]. Hmm. Curious. p[sup]2[/sup] (for the system) = 0. Do you see the kinetic energy in that case as being 0?
Perhaps you should demonstrate the full derivation of “E[sup]2[/sup] = m[sup]2[/sup] + p[sup]2[/sup].”
I assume you are deriving it from “E[sub]t[/sub][sup]2[/sup] = E[sub]k[/sub][sup]2[/sup] + E[sub]r[/sub][sup]2[/sup]”. If you don’t feel that E[sub]k[/sub] = 0 for my example, please demonstrate how this is not a contradiction.
Just an additional point.
Ring couldn’t possibly have been describing the magnitude of a momentum 4 vector since he used “m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup]” in his argument. Clearly “m” here as derived from the total energy equation is invariant mass. It couldn’t be anything else in that context.
I’m sorry but that is exactly what I was describing. No matter how a closed system internally changes the magnitude of the energy momentum four vector remains invariant.
Energy is the time component and momentum is the space component and depending on the frame these components can vary all over the map. But the magnitude of the four vector always remains the same, and this magnitude is equal to the invariant mass.
Damn, I thought I said I was done talking about this stuff. BTW g8rguy explained it beautifully
The kinetic energy of the system is the sum of the kinetic energies of the individual particles. The way it would work out is as follows:
particle 1: give it momentum p, energy E[sub]1[/sub], and mass m[sub]1[/sub]. We know that these three are not independent, and are tied together by
E[sub]1[/sub][sup]2[/sup] = m[sub]1[/sub][sup]2[/sup] + p[sup]2[/sup]
The kinetic energy for this particle is E[sub]1[/sub] - m[sub]1[/sub], but it’s better to speak of E, m, and p.
particle 2: it has momentum -p, energy E[sub]2[/sub], and mass m[sub]2[/sub]. Again, we also know
E[sub]2[/sub][sup]2[/sup] = m[sub]2[/sub][sup]2[/sup] + p[sup]2[/sup]
For the system as a whole, the energy is just E[sub]1[/sub] + E[sub]2[/sub]. The momentum as a whole is zero. The kinetic energy is E[sub]1[/sub] + E[sub]2[/sub] - m[sub]1[/sub] - m[sub]2[/sub]. But again, kinetic energy isn’t as much of a useful concept in relativity as it is in classical mechanics.
The derivation for E[sup]2[/sup] = m[sup]2[/sup] + p[sup]2[/sup] is actually just that the norm of the momentum four vector is defined to be the invariant mass. They’re the same thing. It comes from the following:
we want a four index velocity-like quantity that transforms appropriately under Lorentz transformations; this is given by the derivative of the position vector by proper time (we can’t use regular time, as it doesn’t transform right). This turns out to be:
u[sup][symbol]m[/symbol][/sup] = [symbol]g[/symbol]*(1,v). Multiplying the entire thing by m gives the momentum four vector.
If you square it out, you’ll find that
p[sup][symbol]m[/symbol][/sup] p[sub][symbol]m[/sub][/symbol] = m[sup]2[/sup] = (p[sup]0[/sup])[sup]2[/sup] - p[sup]2[/sup]. Evaluation of p[sup]0[/sup] in the nonrelativistic limit tells you that p[sup]0[/sup] must be the energy, and hence that we have m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup], as Ring wrote.
catches Ring’s last comment on preview Aww, shucks, but thanks for the compliment!
Thank you for that description. I seem to suddenly be coming down with a fever, so it may take me awhile to respond technically.
So Ring, lets be clear about this. You are saying that you agree that the invariant mass drops to zero in a electron/positron collision?
Or let me put it this way:
If I am reading this correctly I am interpreting Ring to say that the invariant mass is constant here, while g8rguy is saying that the invariant mass drops to 0, as I initially said.
Am I wrong? Sounds like the 2 of you are still expressing different views.
The source of the confusion is this. The invariant mass of the system is constant; I agree entirely with Ring. This is a bit of a confusing concept, at times, however, because we’re used to thinking of mass as being conserved in a different way. I probably shouldn’t have made the comment you quoted, as it’s misleading at best. What I was trying to get at is that you can think of this in two ways.
In one way, you would find the invariant mass for each particle and add that up; this is NOT constant; at the beginning of my thought experiment it was 2m, but at the end it was 0. It does, however, agree more with the way we think about mass classically.
In the second, which is really the right way to do it, you would add the two momentum vectors and find the invariant mass of the system as a whole; this IS constant, but it doesn’t make much sense when we think about it as we’re used to doing. Nevertheless, it’s also perfectly correct.
In other words (darn it, I should have phrased it this way to begin with), the total mass M is the same initially and finally. Initially, part of that (2m) is from the mass of the individual particles. At the end, none of it is due to the mass of the individual particles. But it’s the same throughout. It’s not at all intuitive, but that’s the way to think about it.
Okay, g8rguy let me ask this. Do you assert that the inertial and gravitational properties of the system remain constant before and after an antimatter collision?
Well I’m not g8rguybut the answer is yes. And this is one of the reasons I think this concept is important. A lot of people associate mass with matter, but mass doesn’t necessarily have anything to do with matter. It’s a property of a system, and the system could be entirely composed of radiation. In fact if the momentum of a system of photons is zero then m = E and the system of photons would gravitate exactly like a mass with a mass of m.
Yes, they should, although I’m not sure this is all that well defined; how do you measure the intertial and gravitational properties of a pair of photons that are speeding away from each other? But if you had the system in a box, the box would behave the same before and after the collision. In both cases, the system would have no momentum, but still energy, and hence would create the same gravitational effects, modulo differences in the distribution of energy.
In practice, of course, putting the system in a box isn’t possible, and the entire thing will be complicated enormously by the fact that the system won’t be closed. The photons will interact differently with the outside world than will the electrons and positrons, and immediately you’ll run into difficulties. This is not a function of the mass of the system per se, merely of its makeup.
g8rguyI like to use the example that a nuclear explosion wholly contained in a vault will weigh the same both before and after the detonation.
Undead DudeYou might find this sci.physics post by Dr. Steve Carlip interesting
"Suppose we are at the center of an in coming spherical shell of light, which will converge on Earth in one year. Suppose that it’s very energetic light, so energetic that its Schwarzschild radius is half a light year. Note that while this scenarios highly unlikely, it can’t be ruled out by observation, since if it’s true, we won’t see this light until it actually reaches us, which time we won’t be in a position to care.
Now, point a flashlight straight up and turn it on. The light from the flashlight will head outward, affected only by the tiny gravitational fields of objects in the Solar System. Half a year from now, though, it will meet the incoming shell of radiation. But at that point, the radiation will be half a light year away—that is, it will be at its Schwarzschild radius. And that means the light won’t be able to escape further; it will be focused back toward the center.
If you think about it, you’ll see that in this scenario, we are at the event horizon of a black hole right now, even though the matter responsible for the black hole is still a light year away. The event horizon is defined as the boundary of the region from which light can’t escape to infinity, and as I’ve just shown, that’s here,'' not out where the ingoing shell of light is. This characteristic is sometimes called the teleological nature of the event horizon (teleology’’ is the study of ``final causes’’— event horizons are not defined by what’s happening locally now, but by what will ultimately happen to light)."
Hmm. That is a very compelling idea, Ring. I’m familiar with the general “sherical shell” concept as it relates to gravitation, so I get the idea, and I see how it could potentially relate to the scenario we were discussing.
I’m gonna play with this in my head for a bit, and review when my illness is gone, but I may be convinced. It will however require me to run more scenarios thru my head when my head is clear, and I may have more questions.